You are given a string s and an integer k.
First, you are allowed to change at most one index in s to another lowercase English letter.
After that, do the following partitioning operation until s is empty:
s containing at most k distinct characters.s and increase the number of partitions by one. The remaining characters (if any) in s maintain their initial order.Return an integer denoting the maximum number of resulting partitions after the operations by optimally choosing at most one index to change.
Example 1:
Input: s = "accca", k = 2
Output: 3
Explanation:
The optimal way is to change s[2] to something other than a and c, for example, b. then it becomes "acbca".
Then we perform the operations:
"ac", we remove it and s becomes "bca"."bc", so we remove it and s becomes "a"."a" and s becomes empty, so the procedure ends.Doing the operations, the string is divided into 3 partitions, so the answer is 3.
Example 2:
Input: s = "aabaab", k = 3
Output: 1
Explanation:
Initially s contains 2 distinct characters, so whichever character we change, it will contain at most 3 distinct characters, so the longest prefix with at most 3 distinct characters would always be all of it, therefore the answer is 1.
Example 3:
Input: s = "xxyz", k = 1
Output: 4
Explanation:
The optimal way is to change s[0] or s[1] to something other than characters in s, for example, to change s[0] to w.
Then s becomes "wxyz", which consists of 4 distinct characters, so as k is 1, it will divide into 4 partitions.
Constraints:
1 <= s.length <= 104s consists only of lowercase English letters.1 <= k <= 26This approach uses recursive backtracking to explore all possible solutions incrementally. At each step, it makes a choice and recursively explores further solutions. If a solution is invalid or not optimal, it backtracks and tries the next option.
The C solution demonstrates a recursive function solve, which repeatedly calls itself reducing the 'step' parameter until it hits the base case. This example outlines the concept of recursive backtracking.
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Time Complexity: O(2^n), as this approach explores all combinations.
Space Complexity: O(n), due to recursive stack space used during backtracking.
This approach uses dynamic programming to store intermediate results of computations, avoiding recalculating them and thus optimizing the time complexity. It typically involves building a table of results initially filled by base cases and then filled iteratively or recursively with subproblem results.
This C solution implements a memoization technique where results for each step are stored in a dp array. This prevents repeated computations.
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Time Complexity: O(n), with each state being computed once.
Space Complexity: O(n), for dp array storage.
| Approach | Complexity |
|---|---|
| Recursive Backtracking | Time Complexity: O(2^n), as this approach explores all combinations. |
| Dynamic Programming | Time Complexity: O(n), with each state being computed once. |
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